If a^2b^2c^3 = 4500. Is b+c = 7 ?

Given: \(a^2*b^2*c^3 = 4500=2^2*3^2*5^3\). Question: is \(b+c=7\)

(1) a, b and c are positive integers --> sure it's possible that \(a=3\), \(b=2\) and \(c=5\) and in this case the answer will be YES but it's also possible that \(a=2\), \(b=3\) and \(c=5\) and in this case the answer will be NO. Not sufficient.

(2) a > b. Clearly insufficient.

(1)+(2) Note that we are not told that a, b and c are prime numbers, so again it's possible that \(a=3>2=b\) and \(c=5\) and in this case the answer will be YES but for example it's also possible that \(a=2^2*3^2=6^2\), \(b=1\) and \(c=5\) (\(a^2*b^2*c^3 =6^2*1^2*5^3=4500\)) and in this case the answer will be NO. Not sufficient.

Answer: E.

Hope it's clear.
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Thank you bunuel. I completely overlooked the part they there is NO mention of then as primes!

Is there any post where you sum up the important such 'overlook-ables' that we must crosscheck before marking a statement Sufficient or Not sufficient.

Thanks in advance.I am in the final leg of my prep,any advise would be much appreciated.

Very often I overlook the possibilty of 0 in equalties. i.e. I check for only +ve and -ve numbers.

A list of the aforementioned "overlookables" would help a lot.

This is a useful post i found.

data-sufficiency-strategy-guide-130264.html#p1132582

Hi,

Sorry I think that answer is B. Here is my opinion: we are just able to re-write 4500=(3^2)*(2^2)*(5^3), right?

just we don't know that which one of 3 or 2 is a & b, correct?

statement 2 is giving us a>b then we know that a=3 and b=2 and problem solved.

I don't know where I went wrong? Can you explain my fault?


Edit: I found it!

Please read the solution: if-a-2b-2c-3-4500-is-b-c-140842.html#p1132262

Pay attention to the examples given:

(1)+(2) Note that we are not told that a, b and c are prime numbers, so again it's possible that \(a=3>2=b\) and \(c=5\) and in this case the answer will be YES but for example it's also possible that \(a=2^2*3^2=6^2\), \(b=1\) and \(c=5\) (\(a^2*b^2*c^3 =6^2*1^2*5^3=4500\)) and in this case the answer will be NO. Not sufficient.

Answer: E.

Hope it's clear.
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yeah, I got it, thanks

If a^2b^2c^3 = 4500. Is b+c = 7 ?
(a^2) (b^2) (c^3) =4500=(30^2) 5
(abc)^2 c= (30^2) 5
If a, b and c are positive integers then we may have c=5 and abc=30 => ab=6

(1) a, b and c are positive integers
a, b and c are positive integers then we may have c=5 and abc=30 => ab=6.
but a=3,b=2 => b+c=7
a=6, b=1 => b+c=6 INSUFFICIENT
(2) a > b
a and b may be fractions and many combinations are posssible for a and b. INSUFFICIENT

combining 1 and 2
a=3,b=2 => b+c=7
a=6, b=1 => b+c=6 INSUFFICIENT

Hence E
Thanks

If \(a^2b^2c^3\) = 4500. Is b+c = 7 ?

\(a^2b^2c^3\) = \(2^23^25^3\)

(1) a, b and c are positive integers

Knowing that a, b and c are positive integers does not tell us which integer is b and which integer is c. Insufficient.

(2) \(a > b\)

There can be many combinations since we're not told that a and b are integers. Insufficient.

(1&2) We still can't determine which integer is b and which integer is c. Insufficient.

Answer is E.
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This is a very smart question. The math is basic but the trap is so easily overlooked.

Posted from my mobile device

The key to solving the problem is breaking it down into a no of possiblities of the number 4500
It can be broken down in the following way :
3^2*2^2*5^3 =6^2*1^2*5^3

Now lets us evaluvate the same in the light of the given statements
a, b and c are positive integers it can be 6 or 7
even with the second statements we are getting both the answers 6 and 7
hence IMO E

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